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I saw this riddle doing the rounds on the internet: https://ed.ted.com/lessons/can-you-solve-the-frog-riddle-derek-abbott

In summary; There is a population of frogs with male:female occurring in 50:50 ratio. There are two patches of ground near you, one containing a single frog, the other containing two frogs. Your survival depends on you finding a female frog in one of these two patches, but you only get to make one attempt. You cannot tell which frogs are which in advance, except that you know that one of the frogs in the patch with two frogs in is male.

The answer given to the riddle is that the odds of the single frog being female is 50%, but the odds of one of the two frogs being female is 2/3 (67%). The explanation being that there are four possible combinations of male female pairs, one is excluded because we know one frog is male, hence 2/3 combinations where we find a female frog in the pair and 1/3 where we don't.

The probabilities just seem wrong to me; can anyone clarify the reason why this is the case?

I suspect that there is a subtly in the framing of the question that I'm missing.

As i read the problem, we have a choice of two options, both of which are simply a 50:50 chance of whether a single frog is male or female. Not knowing which frog in the pair is definitely male should have no effect on the probability of the other.

If I am wrong I really want to understand why!

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    $\begingroup$ Can you restate the riddle here so readers don't have to follow the link (which also may break in the future) and then watch a video? $\endgroup$ – dsaxton Mar 13 '16 at 23:11
  • $\begingroup$ It seems to me that one has to make strong assumptions in order to obtain any answer. E.g., supposing male frogs croak only in the presence of a female, you would obtain one answer; but supposing that they tend to croak in the presence of another male, you would obtain a different answer (and make a different decision). Or what if females are not gregarious and tend to avoid other frogs? You would make yet a third decision. Although it's clearly intended that you ignore all such considerations, contemplating them may help you understand why the odds you compute are not necessarily 50:50. $\endgroup$ – whuber Mar 15 '16 at 22:43
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    $\begingroup$ The TED-Ed frog riddle answer is wrong. There is a very detailed answer here: duckware.com/tedfrog $\endgroup$ – Tim Greenberg Jan 26 '17 at 16:54
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Let's look at the pair of frogs. Male frogs are identified by croaking in the video.

As explained in the video, before we hear any croaking, there are 4 equally likely outcomes given 2 frogs:

  • Frog 1 is Male, Frog 2 is Male
  • Frog 1 is Female, Frog 2 is Male
  • Frog 1 is Male, Frog 2 is Female
  • Frog 1 is Female, Frog 2 is Female

Making the assumptions about males and females occurring equally and independently, our sample space is $\{(M,M),(F,M),(M,F),(F,F)\}$, and we have probability $1/4$ for each element.

Now, once we hear the croak coming from this pair, we know that at least one frog is male. Thus the event $(F,F)$ is impossible. We then have a new, reduced sample space induced by this condition: $\{(M,M),(F,M),(M,F)\}$. Each remaining possibility is still equally likely, and the probability of all the events added together must be $1$. So the probability of each of these three events in the new sample space must be $1/3$.

The only event that ends badly for us is $(M,M)$, so there is a $2/3$ chance of survival.


More formally, the definition of conditional probability says:

$$P(A|B) = \frac{P(A \cap B)}{P(B)}$$ So if $A$ is the event that at least one female is present and $B$ is the event that at least one male is present, we have: \begin{align}P(\text{F given at least 1 M}) &= \frac{P(\text{F and at least 1 male})}{P(\text{at least 1 M})}\\ &= \frac{P(\text{1 M and 1 F})}{P(\text{1 M or 2 M})} \\ &= \frac{P[(M,F),(F,M)]}{P[(M,M),(F,M),(M,F)]} \\ &= \frac{1/2}{3/4} = 2/3 \end{align}

This is really the same procedure we reasoned through as above.

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  • $\begingroup$ Hi mb7744, thanks for the quick response. I understand the answer as laid out, however this does look to me like double counting which is why I'm struggling to accept the answer. (M,F)=(F,M), surely, and if not, why? $\endgroup$ – Jernau Mar 13 '16 at 23:50
  • $\begingroup$ (M,F) and (F,M) are not the same event. If one frog is named Alex and the other frog is named Taylor, Alex could be the female and Taylor the male OR vice versa. Alex and Taylor would probably disagree that this distinction is meaningless. Now, you could view the two events as equivalent. However, then your three outcomes (M,M),(F,F) and (M,F) are not equally likely. The mixed pairing is twice as likely. This is the same reason that you are much more likely to roll a 7 on a pair of dice than a 2, even if you view all of the different ways of rolling 7 as equivalent. $\endgroup$ – mb7744 Mar 13 '16 at 23:57
  • $\begingroup$ Hi, I think this helps clarify where i'm not 'getting' the riddle. If I may restate the problem as I'm seeing it, replace frog with a coin toss (or a dice roll). If you got to flip two coins and exclude certain combinations I would completely accept the answer. In the riddle's analogy however, i read this as we only get one coin toss. The other has already been made and cannot change the outcome of the other. Not knowing which of the two outcome has already been determined doesn't allow us to flip two coins and choose which outcomes to include or exclude. So using the dice roll analogy..... $\endgroup$ – Jernau Mar 14 '16 at 6:57
  • $\begingroup$ ...you get to roll two dice, but unknown to you one dice' outcome has already been decided. You only have 1/6 chance of making any number 7-12. Am I wrong here? $\endgroup$ – Jernau Mar 14 '16 at 6:57
  • $\begingroup$ If we look at all the pairs of equally-likely outcomes in dice rolling, order matters. Imagine one die is blue and the other red, and we write our outcomes with the blue die first and the red die last. Then the outcome (1,2) is not the same as the outcome (2,1). And, as before, the probability of rolling a "1 and a 2, regardless of order" will be twice as much as, say, rolling a pair of 2s. For your last question, I am assuming you meant to say one die's outcome was decided to be 6. In that case you are correct. $\endgroup$ – mb7744 Mar 14 '16 at 12:50
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Since the math is already laid out I'll try to provide some intuition. The issue is that knowing that at least one frog is male is different from knowing that any particular frog is male. The former case carries less information and this effectively increases our chances over the latter situation.

Call the frogs left and right, and suppose we are told that the right frog is male. Then we have eliminated two possible events from the sample space: the event where both frogs are female and the event where the left frog is male and the right frog is female. Now the probability truly is one half and it doesn't matter which one we choose. The exact same argument is true if we learn that the left frog is male.

But if we are told only that at least one frog is male, which is what happens when we hear the croak, then we cannot eliminate the event that the left frog is male and the right frog is female. We can only eliminate the event that both are female, which makes the event that at least one is female more likely than the previous setting.

I think the reason why this is confusing is that we naturally think learning that at least one is male should make us disinclined to choose the pair of frogs. It is true that this information makes it less probable that at least one is female, but recognize also that there was a full three quarters chance of at least one female before we learned anything at all. It's the ambiguity of the information we receive which makes it so we should still prefer the two frogs over the one.

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  • $\begingroup$ Thanks dsaxton, intuitively I opted for the two frogs, but my reasoning told me either choice was equally probable. $\endgroup$ – Jernau Mar 14 '16 at 7:02
  • $\begingroup$ Thanks dsaxton, I suspect it's the phrasing of the riddle that is throwing me. As encountered, the two frogs are not distinguishable (without further information), so I am not seeing the (M,F), (F,M) distinction as meaningful in this context. I am not convinced that my reasoning is faulty, but my apologies if I am just being a bit slow. $\endgroup$ – Jernau Mar 14 '16 at 7:11
  • $\begingroup$ Thanks again dsaxton. As mentioned above, I've found the mental hang up I was having and can see now why the answer is the right answer (and the question I was actually trying to answer). Thanks again for your help, seeing the answer is just not the same as having the help to really understand it. $\endgroup$ – Jernau Mar 14 '16 at 20:10
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Your intuition is correct in this case. As the problem is stated your odds of survival are 50%. The video incorrectly states the problem space based on the information we have and therefore comes to an incorrect conclusion. The correct problem space contains 8 conditions and is as follows.

We have two frogs on a log, and one of them has croaked what are our possibilities? (M designates male, F designates female and c designates croaked, first position is left, second position is right)

[
  [Mc, M], 
  [M, Mc],
  [Mc, F], 
  [M, Fc], (X No Male croak) 
  [Fc, M], (X No Male croak)
  [F, Mc], 
  [Fc, F], (X No Male croak)
  [F, Fc], (X No Male croak)
]

Each case is equally likely based on the information that we have, when we eliminate the conditions given the knowledge that a male frog has croaked. We find that there are 4 outcomes to expect. Left male frog croaked next to a right male frog that was silent. Right male frog croaked next to a left male frog that was silent. Or there was a croaking male frog paired with a single female frog in either direction. For an intuitive way to understand this, the two male frogs are twice as likely to croak than the single male frog paired with a female, so we have to weight it appropriately.

You could also divide the search space by croaking frog (C) and non croaking frog (N). Since the croaking frog is 100% a male, you can eliminate it from your search since it has no chance of helping you survive. While the author intended to create a "monty hall problem" they inadvertently created a "boy or girl paradox".

The following questions yield different results:

Given that there is a male what is the likelihood the other is female?

Given that a male frog croaked what is the likelihood the other is female?

I know more information in the second case

https://en.wikipedia.org/wiki/Monty_Hall_problem

https://en.wikipedia.org/wiki/Boy_or_Girl_paradox

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A clearer answer to this, since the previous was too long and not easy to understand.

The possible outcomes are different, although I used same letters. To make clear the sample space, I will describe the possible outcomes

M M --> "The male is on the left" - "A random male on the right"

M F --> "The male is on the left" - "A random female on the right"

M M --> "The male is on the right" - "A random male on the left"

M F --> "The male is on the right" - "A random female on the left"

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  • $\begingroup$ You are double counting the MM case. You can't just enumerate all the possible scenarios without taking into account whether you're arriving at the same scenario through different paths. $\endgroup$ – Acccumulation Jan 8 '19 at 23:20
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The problem I have with this problem, is that the solution seems to be using different rules for what it considers a possible result for the two frogs being male and female, and male and male.

The F/M pair, and the M/F pair, are different because we don't know whether the first frog or the second frog is male, so F/M and M/F are two separate possibilities, even though the result still amounts to "one female frog, one male frog".

But the M/M pair is only considered one possible result, even though the same logic should apply: we don't know which frog is the one that made the croaking sound, so either frog could be the one we heard, and the other one could still be male, it just didn't happen to croak.

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  • $\begingroup$ This is more in the nature of a comment than an answer to the "riddle." Please change it to a comment and delete this "answer." $\endgroup$ – Mike Hunter Apr 1 '16 at 11:32
  • $\begingroup$ @DJohnson Actually, this is an answer to the riddle, although the later answer from tomciopp explains it more clearly. $\endgroup$ – Acccumulation Jan 8 '19 at 23:18
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Not knowing anything: $\{(M,M), (M,F), (F,M), (F,F)\}$. Three pairs with at least one female out of four possible combinations: $3/4$ or $75\%$

Knowing the first one is male: $\{(M,M), (M,F)\}$. One pair with at least one female out of two possible combinations: $1/2$ or $50\%$

Knowing that there is at least one male: $\{(M,M), (M,F), (F,M)\}$. Two pairs with at least one female out of three possible combinations: $2/3$ or $67\%$

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Before we hear any croaking, there are 4 equally likely outcomes given 2 frogs:

Frog 1 is Male, Frog 2 is Male

Frog 1 is Female, Frog 2 is Male

Frog 1 is Male, Frog 2 is Female

Frog 1 is Female, Frog 2 is Female

Making the assumptions about males and females occurring equally and independently, our sample space is {(M,M),(F,M),(M,F),(F,F)}, and we have probability 1/4 for each element.

Once we hear the croak coming from this pair, we know that at least one frog is male. This male can equally likely be Frog 1 or Frog 2. So there are 2 equally likely outcomes for the Frog 1:

Frog 1 is Male

Frog 1 is Random Frog

Making the assumptions about males and females occurring equally and independently, the Random Frog is equally likely to be a Random Male or a Random Female.

P(Frog 1 is Random Male given Frog 1 is Random Frog)=P(Frog 1 is Random Female given Frog 1 is Random Frog)=1/2

P(Frog 1 is Random Male and Frog 1 is Random Frog)=P(Frog 1 is Random Frog)P(Frog 1 is Random Male given Frog 1 is Random Frog)=(1/2)(1/2)=1/4

P(Frog 1 is Random Female and Frog 1 is Random Frog)=P(Frog 1 is Random Frog)P(Frog 1 is Random Female given Frog 1 is Random Frog)=(1/2)(1/2)=1/4

So there are 3 possible outcomes for the Frog 1:

Frog 1 is Male

Frog 1 is Random Male

Frog 1 is Random Female

and probabilities are:

P(Frog 1 is Male)=1/2

P(Frog 1 is Random Male)=1/4

P(Frog 1 is Random Female)=1/4

Now, for each possible outcome for Frog 1, there are 2 possible outcomes for the Frog 2:

Frog 2 is Male

Frog 2 is Random Frog

For each possible outcome for Frog 1, the Random Frog is equally likely to be a Random Male or a Random Female.

So, for each possible outcome for Frog 1, there are 3 possible outcomes for the Frog 2:

Frog 2 is Male

Frog 2 is Random Male

Frog 2 is Random Female

P(Frog 2 is Male given Frog 1 is Male)=0

P(Frog 2 is Male given Frog 1 is Random Male)=1

P(Frog 2 is Male given Frog 1 is Random Female)=1

P(Frog 2 is Random Male given Frog 1 is Male)=1/2

P(Frog 2 is Random Male given Frog 1 is Random Male)=0

P(Frog 2 is Random Male given Frog 1 is Random Female)=0

P(Frog 2 is Random Female given Frog 1 is Male)=1/2

P(Frog 2 is Random Female given Frog 1 is Random Male)=0

P(Frog 2 is Random Female given Frog 1 is Random Female)=0

P(Frog 2 is Random Male and Frog 1 is Male)=P(Frog 1 is Male)P(Frog 2 is Random Male given Frog 1 is Male)=(1/2)(1/2)=1/4

P(Frog 2 is Random Female and Frog 1 is Male)=P(Frog 1 is Male)P(Frog 2 is Random Female given Frog 1 is Male)=(1/2)(1/2)=1/4

P(Frog 2 is Male and Frog 1 is Random Male)=P(Frog 1 is Random Male)*P(Frog 2 is Male given Frog 1 is Random Male)=(1/4)*1=1/4

P(Frog 2 is Male and Frog 1 is Random Female)=P(Frog 1 is Random Female)*P(Frog 2 is Male given Frog 1 is Random Female)=(1/4)*1=1/4

So, our sample space is {(Male,Random Male),(Male,Random Female),(Random Male,Male),(Random Female,Male)}, and we have probability 1/4 for each element.

P(F given at least 1 M)=P(F and at least 1 male)/P(at least 1 M)=P(1 M and 1 F)/P(1 M or 2 M)=P[(Male,Random Female),(Random Female,Male)]/P[(Male,Random Male),(Male,Random Female),(Random Male,Male),(Random Female,Male)]=(1/2)/(4/4)=1/2

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    $\begingroup$ Did you copy and paste from my answer and remove the formatting? $\endgroup$ – mb7744 Apr 1 '16 at 13:01
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    $\begingroup$ Well, first of all, copying and pasting a part of someone else's answer without even mentioning it is unacceptable. That aside, if you think that you have reached a different result, is there a more concise way for you to explain it? You have written a lot of disconnected equations without any explanation. $\endgroup$ – mb7744 Apr 2 '16 at 4:16
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    $\begingroup$ It's not literature but it is still rude. Now, with regards to your answer versus mine: I find yours nonsensical. What is the meaning of the outcome "Frog 2 is Random Frog"? $\endgroup$ – mb7744 Apr 2 '16 at 17:31
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    $\begingroup$ Your answer was the only one calculating conditional probabilities. Using same terms could help comparing and see which part is the same and which is different. I may say, I find other answers nonsensical too, but I didnt say so because it would be rude ;) . If you dont understand sth, you can jusk ask for clarifications. "Frog 2 is Random Frog" means it is not the male frog known to be in the pair.... $\endgroup$ – pit847 Apr 2 '16 at 18:52
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    $\begingroup$ There are two sources of randomness, one coming from the male frog known to be in the pair, the other coming from the frog population. Since we know the male frog is there, the uncertainty is just about the position. Is it frog 1 or frog 2? Or, is it on the left or on the right? My advice is, use tree diagram to build sample space from scratch and use all info available. $\endgroup$ – pit847 Apr 2 '16 at 18:52

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